Some Basic Concepts in Chemistry MCQs with Answers – Part 5 (NEET Chemistry)

Explanation: \(\textbf{Interpreting the hydrate formula:}\) Every original sodium carbonate formula unit is associated with ten water molecules. The problem states that exactly half the original water of crystallisation escapes. The sodium carbonate constituent remains unchanged. \(\textbf{Determining the original water amount:}\) For \(0.20\,\mathrm{mol}\) of hydrate formula units, \[ n_{\mathrm{H_2O,initial}}=10(0.20) \] \[ n_{\mathrm{H_2O,initial}}=2.00\,\mathrm{mol} \] \(\textbf{Calculating the amount lost:}\) \[ n_{\mathrm{H_2O,lost}}=\frac12(2.00) \] \[ n_{\mathrm{H_2O,lost}}=1.00\,\mathrm{mol} \] The remaining water amount is \[ n_{\mathrm{H_2O,remaining}}=2.00-1.00 \] \[ n_{\mathrm{H_2O,remaining}}=1.00\,\mathrm{mol} \] \(\textbf{Recovering water per carbonate formula unit:}\) The sodium carbonate amount remains \(0.20\,\mathrm{mol}\). Therefore, \[ x=\frac{n_{\mathrm{H_2O,remaining}}}{n_{\mathrm{Na_2CO_3}}} \] \[ x=\frac{1.00}{0.20} \] \[ x=5 \] The retained material has the stipulated composition \(\mathrm{Na_2CO_3\cdot5H_2O}\). \(\textbf{Calculating the original sample mass:}\) The anhydrous sodium carbonate molar mass is \[ M_{\mathrm{Na_2CO_3}}=2(23)+12+3(16) \] \[ M_{\mathrm{Na_2CO_3}}=106\,\mathrm{g\,mol^{-1}} \] The original hydrate molar mass is \[ M_{\mathrm{original}}=106+10(18) \] \[ M_{\mathrm{original}}=286\,\mathrm{g\,mol^{-1}} \] Therefore, \[ m_{\mathrm{initial}}=0.20(286) \] \[ m_{\mathrm{initial}}=57.2\,\text{g} \] \(\textbf{Calculating the retained solid mass:}\) The sodium carbonate constituent weighs \[ m_{\mathrm{Na_2CO_3}}=0.20(106) \] \[ m_{\mathrm{Na_2CO_3}}=21.2\,\text{g} \] The retained water weighs \[ m_{\mathrm{H_2O,retained}}=1.00(18) \] \[ m_{\mathrm{H_2O,retained}}=18.0\,\text{g} \] The final solid mass is \[ m_{\mathrm{solid}}=21.2+18.0 \] \[ m_{\mathrm{solid}}=39.2\,\text{g} \] \(\textbf{Finding the percentage remaining:}\) \[ \%\text{solid remaining}=\frac{39.2}{57.2}\times100 \] \[ \%\text{solid remaining}\approx68.53\% \] \(\textbf{Conservation check:}\) The escaped water weighs \(1.00(18)=18.0\) g. The retained solid and escaped water together weigh \(39.2+18.0=57.2\) g. The calculation assumes the stated uniform retained-hydrate composition; it does not rely on an unspecified real-world dehydration pathway. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{Na_2CO_3\cdot5H_2O},\quad\%\text{solid remaining}=68.53\%} \]

Explanation: \(\textbf{Interpreting the ionic formula:}\) The compound contains two M ions and three sulphate ions per formula unit. Complete dissociation therefore gives five independently dispersed ions per formula unit. For \(0.050\,\mathrm{mol}\) formula units, \[ n_{\mathrm{ions}}=5(0.050) \] \[ n_{\mathrm{ions}}=0.250\,\mathrm{mol} \] This agrees with the observed ion count. \(\textbf{Recovering the ionic charge through neutrality:}\) Three sulphate ions each carry charge \(2-\). Their combined negative charge per formula unit is \(6-\). The two M ions must together supply \(6+\). Therefore, each M ion has charge \(3+\). The metal ion is trivalent. \(\textbf{Determining the compound's molar mass:}\) The measured sample contains \(0.050\,\mathrm{mol}\) formula units and weighs 17.1 g. \[ M_{\mathrm{compound}}=\frac{17.1}{0.050} \] \[ M_{\mathrm{compound}}=342\,\mathrm{g\,mol^{-1}} \] \(\textbf{Removing the sulphate contribution:}\) One sulphate group has formula mass \[ m_{\mathrm{SO_4}}=32+4(16) \] \[ m_{\mathrm{SO_4}}=96\,\mathrm{u} \] Three groups contribute \[ 3(96)=288\,\mathrm{u} \] Let the atomic mass of M be \(a\,\mathrm{u}\). \[ 2a+288=342 \] \[ 2a=54 \] \[ a=27 \] Thus, the atomic mass is \(27\,\mathrm{u}\). \(\textbf{Checking the independence of the observations:}\) The ion-count observation confirms the number of ions represented by the formula but does not establish the atomic mass. Electrical neutrality fixes the charge of M. The sample mass and formula-unit amount provide the molar mass needed to determine its atomic mass. These are different types of information, and combining them yields the complete conclusion. The correct result is a trivalent metal ion with atomic mass \(27\,\mathrm{u}\).

Explanation: \(\textbf{Verifying the stated compound molar mass:}\) The formula contains two X atoms and three Y atoms per formula unit. Using the supplied average atomic masses, \[ M_{\mathrm{X_2Y_3}}=2(56)+3(16) \] \[ M_{\mathrm{X_2Y_3}}=112+48 \] \[ M_{\mathrm{X_2Y_3}}=160\,\mathrm{g\,mol^{-1}} \] The measured molar mass therefore agrees with the chemical formula. \(\textbf{Recovering the amount of formula units:}\) For the separate 8.0 g sample, \[ n_{\mathrm{formula\ units}}=\frac{8.0}{160} \] \[ n_{\mathrm{formula\ units}}=0.050\,\mathrm{mol} \] \(\textbf{Counting the Y atoms:}\) Every formula unit contains three Y atoms. \[ n_Y=3n_{\mathrm{formula\ units}} \] \[ n_Y=3(0.050) \] \[ n_Y=0.150\,\mathrm{mol} \] The researcher's proposed amount is correct. \(\textbf{Checking the mass contribution of Y:}\) The Y atoms contribute \[ m_Y=0.150(16) \] \[ m_Y=2.4\,\text{g} \] The X atoms contribute \[ n_X=2(0.050)=0.100\,\mathrm{mol} \] \[ m_X=0.100(56)=5.6\,\text{g} \] Together, \[ m_X+m_Y=5.6+2.4 \] \[ m_X+m_Y=8.0\,\text{g} \] This reproduces the sample mass. The distinction is between the amount of formula units and the amount of constituent atoms represented by those formula units. A sample containing \(0.050\,\mathrm{mol}\) of \(\mathrm{X_2Y_3}\) does not contain merely \(0.050\,\mathrm{mol}\) of Y atoms. The external subscript three makes their amount three times greater. Option A is the unique correct evaluation.

Explanation: \(\textbf{Recovering the initial sulphur oxide composition:}\) Let \(x\) and \(y\) be the amounts of \(\mathrm{SO_2}\) and \(\mathrm{SO_3}\), respectively. The total molecular amount gives \[ x+y=0.50 \qquad (1) \] Their molar masses are \[ M_{\mathrm{SO_2}}=64\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{SO_3}}=80\,\mathrm{g\,mol^{-1}} \] The original mixture mass gives \[ 64x+80y=36.8 \qquad (2) \] \(\textbf{Solving the original mixture:}\) From Equation (1), \[ x=0.50-y \] Substituting, \[ 64(0.50-y)+80y=36.8 \] \[ 32.0+16y=36.8 \] \[ 16y=4.8 \] \[ y=0.30\,\mathrm{mol} \] Therefore, \[ x=0.50-0.30=0.20\,\mathrm{mol} \] \(\textbf{Applying helium addition:}\) The added helium amount is \(0.10\,\mathrm{mol}\). No reaction occurs and no sulphur oxide is removed. The new total molecular amount is \[ n_{\mathrm{total,new}}=0.50+0.10 \] \[ n_{\mathrm{total,new}}=0.60\,\mathrm{mol} \] The sulphur trioxide amount remains \(0.30\,\mathrm{mol}\). Its new mole fraction is \[ X_{\mathrm{SO_3,new}}=\frac{0.30}{0.60} \] \[ X_{\mathrm{SO_3,new}}=0.50 \] \(\textbf{Evaluating the mass fraction:}\) The sulphur trioxide mass is \[ m_{\mathrm{SO_3}}=0.30(80) \] \[ m_{\mathrm{SO_3}}=24.0\,\text{g} \] The helium contributes \[ m_{\mathrm{He}}=0.10(4)=0.4\,\text{g} \] The new mixture mass is \[ m_{\mathrm{new}}=36.8+0.4 \] \[ m_{\mathrm{new}}=37.2\,\text{g} \] The sulphur trioxide mass fraction changes from \[ w_{\mathrm{SO_3,initial}}=\frac{24.0}{36.8} \] to \[ w_{\mathrm{SO_3,new}}=\frac{24.0}{37.2} \] The second fraction is smaller. \(\textbf{Checking the options:}\) The calculated sulphur trioxide mole fraction is 0.50. Adding inert helium increases both the total molecular amount and the total mass of the mixture. The new sulphur trioxide mole fraction is 0.50, and its mass fraction decreases because the sulphur trioxide mass is unchanged while the denominator increases. Both changes arise from the stated helium addition without a chemical reaction. \(\textbf{Final answer:}\) \[ \boxed{X_{\mathrm{SO_3,new}}=0.50,\quad w_{\mathrm{SO_3}}\text{ decreases}} \]

Explanation: \(\textbf{Recovering the original molecular populations:}\) Let \(x\) be the amount of oxygen molecules and \(y\) the amount of ozone molecules. The total molecular amount gives \[ x+y=0.50 \qquad (1) \] \(\textbf{Using the conserved oxygen-atom inventory:}\) Each oxygen molecule contains two oxygen atoms. Each ozone molecule contains three. Therefore, \[ 2x+3y=1.20 \qquad (2) \] \(\textbf{Solving the mixture composition:}\) Multiplying Equation (1) by two, \[ 2x+2y=1.00 \] Subtracting this equation from Equation (2), \[ y=1.20-1.00 \] \[ y=0.20\,\mathrm{mol} \] Therefore, \[ x=0.50-0.20 \] \[ x=0.30\,\mathrm{mol} \] \(\textbf{Converting all ozone into oxygen:}\) Two moles of ozone produce three moles of oxygen molecules. The new oxygen formed is \[ n_{\mathrm{O_2,new}}=0.20\left(\frac32\right) \] \[ n_{\mathrm{O_2,new}}=0.30\,\mathrm{mol} \] \(\textbf{Determining the final molecular amount:}\) The initial oxygen molecules remain present. The final oxygen amount is therefore \[ n_{\mathrm{O_2,final}}=0.30+0.30 \] \[ n_{\mathrm{O_2,final}}=0.60\,\mathrm{mol} \] All ozone has been converted, so the total final gas amount is also \(0.60\,\mathrm{mol}\). \(\textbf{Calculating the pressure ratio:}\) At constant temperature and volume, \[ \frac{P_f}{P_i}=\frac{n_f}{n_i} \] Substituting, \[ \frac{P_f}{P_i}=\frac{0.60}{0.50} \] \[ \frac{P_f}{P_i}=1.20 \] \(\textbf{Independent atom-count check:}\) The final oxygen molecules contain \[ n_{O,\mathrm{final}}=2(0.60) \] \[ n_{O,\mathrm{final}}=1.20\,\mathrm{mol} \] This equals the original oxygen-atom amount. The molecular count increases during ozone decomposition, but the total number of oxygen atoms and the gas mass remain unchanged. The pressure increase follows from the larger number of gaseous particles, not from creation of matter. \(\textbf{Final answer:}\) \[ \boxed{\frac{P_f}{P_i}=1.20,\quad n_{\mathrm{O_2,final}}=0.60\,\mathrm{mol}} \]

Explanation: \(\textbf{Interpreting the pressure measurement:}\) Only gaseous species contribute to the ideal-gas pressure. The methanol product is specified to remain liquid, so its molecular amount must not be included in the final gaseous inventory. Initially, \[ n_{\mathrm{gas},i}=0.30+0.80+0.20 \] \[ n_{\mathrm{gas},i}=1.30\,\mathrm{mol} \] At constant temperature and volume, \[ \frac{P_f}{P_i}=\frac{n_{\mathrm{gas},f}}{n_{\mathrm{gas},i}} \] Thus, \[ n_{\mathrm{gas},f}=\frac7{13}(1.30) \] \[ n_{\mathrm{gas},f}=0.70\,\mathrm{mol} \] \(\textbf{Connecting molecular disappearance with reaction extent:}\) Let \(x\) be the amount of methanol formed. For each mole of methanol produced, one mole of CO and two moles of hydrogen disappear from the gas phase. Therefore, \[ n_{\mathrm{CO},f}=0.30-x \] \[ n_{\mathrm{H_2},f}=0.80-2x \] Helium remains unchanged at \(0.20\,\mathrm{mol}\). The final gaseous amount is \[ n_{\mathrm{gas},f}=(0.30-x)+(0.80-2x)+0.20 \] \[ n_{\mathrm{gas},f}=1.30-3x \] Using the pressure-derived value, \[ 0.70=1.30-3x \] \[ 3x=0.60 \] \[ x=0.20\,\mathrm{mol} \] \(\textbf{Determining the liquid product mass:}\) \[ M_{\mathrm{CH_3OH}}=12+4(1)+16 \] \[ M_{\mathrm{CH_3OH}}=32\,\mathrm{g\,mol^{-1}} \] Therefore, \[ m_{\mathrm{CH_3OH}}=0.20(32) \] \[ m_{\mathrm{CH_3OH}}=6.4\,\text{g} \] \(\textbf{Reconstructing the remaining gases:}\) \[ n_{\mathrm{CO},f}=0.30-0.20=0.10\,\mathrm{mol} \] \[ n_{\mathrm{H_2},f}=0.80-2(0.20) \] \[ n_{\mathrm{H_2},f}=0.40\,\mathrm{mol} \] The total is \[ 0.10+0.40+0.20=0.70\,\mathrm{mol} \] The hydrogen mole fraction is \[ X_{\mathrm{H_2}}=\frac{0.40}{0.70} \] \[ X_{\mathrm{H_2}}=\frac47 \] \(\textbf{Final consistency check:}\) The derived hydrogen mole fraction is \(\frac47\), not \(\frac13\) or \(\frac49\). The methanol mass is 6.4 g, and the recovered gas inventory is 0.10 mol CO, 0.40 mol hydrogen and 0.20 mol helium. The pressure observation uniquely determines the reaction extent, and the final gaseous composition follows directly from it. The two requested quantities are supported independently by product stoichiometry and the reconstructed gas amounts. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{CH_3OH}}=6.4\,\text{g},\quad X_{\mathrm{H_2}}=\frac47} \]

Explanation: \(\textbf{Recovering the two combustible-gas amounts:}\) Let \(x\) be the amount of methane and \(y\) the amount of ethane. The original molecular amount gives \[ x+y=0.40 \qquad (1) \] Methane produces one mole of carbon dioxide per mole consumed. Ethane produces two moles of carbon dioxide per mole consumed. The measured product amount therefore gives \[ x+2y=0.60 \qquad (2) \] \(\textbf{Resolving the mixture composition:}\) Subtracting Equation (1) from Equation (2), \[ y=0.60-0.40 \] \[ y=0.20\,\mathrm{mol} \] Consequently, \[ x=0.40-0.20 \] \[ x=0.20\,\mathrm{mol} \] \(\textbf{Calculating the starting mixture mass:}\) The methane molar mass is \[ M_{\mathrm{CH_4}}=12+4(1)=16\,\mathrm{g\,mol^{-1}} \] The ethane molar mass is \[ M_{\mathrm{C_2H_6}}=2(12)+6(1)=30\,\mathrm{g\,mol^{-1}} \] Thus, \[ m_{\mathrm{mixture}}=0.20(16)+0.20(30) \] \[ m_{\mathrm{mixture}}=3.2+6.0 \] \[ m_{\mathrm{mixture}}=9.2\,\text{g} \] \(\textbf{Determining the oxygen required by each fuel:}\) One mole of methane requires two moles of oxygen. \[ n_{\mathrm{O_2,CH_4}}=2(0.20) \] \[ n_{\mathrm{O_2,CH_4}}=0.40\,\mathrm{mol} \] One mole of ethane requires \(3.5\) moles of oxygen. \[ n_{\mathrm{O_2,C_2H_6}}=\frac72(0.20) \] \[ n_{\mathrm{O_2,C_2H_6}}=0.70\,\mathrm{mol} \] The total oxygen consumption is \[ n_{\mathrm{O_2,total}}=0.40+0.70 \] \[ n_{\mathrm{O_2,total}}=1.10\,\mathrm{mol} \] \(\textbf{Checking the hydrogen inventory through water formation:}\) Methane produces \[ n_{\mathrm{H_2O,CH_4}}=2(0.20)=0.40\,\mathrm{mol} \] Ethane produces \[ n_{\mathrm{H_2O,C_2H_6}}=3(0.20)=0.60\,\mathrm{mol} \] Therefore, \[ n_{\mathrm{H_2O,total}}=1.00\,\mathrm{mol} \] \(\textbf{Mass-conservation verification:}\) The reactants weigh \[ m_{\mathrm{reactants}}=9.2+1.10(32) \] \[ m_{\mathrm{reactants}}=44.4\,\text{g} \] The products weigh \[ m_{\mathrm{products}}=0.60(44)+1.00(18) \] \[ m_{\mathrm{products}}=26.4+18.0 \] \[ m_{\mathrm{products}}=44.4\,\text{g} \] The mass balance agrees. The carbon dioxide amount determines the fuel composition because methane and ethane contain different numbers of carbon atoms per molecule. That composition must then be used to calculate the combined oxygen requirement. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{mixture}}=9.2\,\text{g},\quad n_{\mathrm{O_2}}=1.10\,\mathrm{mol}} \]

Explanation: \(\textbf{Using hydrogen to isolate the magnesium-metal contribution:}\) Only elemental magnesium produces hydrogen in the specified reactions. Magnesium oxide reacts with the acid without producing hydrogen gas. The magnesium-to-hydrogen mole ratio is \(1:1\). Therefore, \[ n_{\mathrm{Mg}}=n_{\mathrm{H_2}} \] \[ n_{\mathrm{Mg}}=0.20\,\mathrm{mol} \] The original magnesium-metal mass is \[ m_{\mathrm{Mg}}=0.20(24) \] \[ m_{\mathrm{Mg}}=4.8\,\text{g} \] \(\textbf{Recovering the magnesium oxide contribution:}\) The mixture weighs 8.0 g. Thus, \[ m_{\mathrm{MgO}}=8.0-4.8 \] \[ m_{\mathrm{MgO}}=3.2\,\text{g} \] The molar mass of magnesium oxide is \[ M_{\mathrm{MgO}}=24+16 \] \[ M_{\mathrm{MgO}}=40\,\mathrm{g\,mol^{-1}} \] Therefore, \[ n_{\mathrm{MgO}}=\frac{3.2}{40} \] \[ n_{\mathrm{MgO}}=0.08\,\mathrm{mol} \] \(\textbf{Calculating the oxide mass percentage:}\) \[ \%\mathrm{MgO}=\frac{3.2}{8.0}\times100 \] \[ \%\mathrm{MgO}=40\% \] \(\textbf{Counting acid consumption separately:}\) Every mole of magnesium metal consumes two moles of HCl. \[ n_{\mathrm{HCl,Mg}}=2(0.20) \] \[ n_{\mathrm{HCl,Mg}}=0.40\,\mathrm{mol} \] Every mole of magnesium oxide also consumes two moles of HCl. \[ n_{\mathrm{HCl,MgO}}=2(0.08) \] \[ n_{\mathrm{HCl,MgO}}=0.16\,\mathrm{mol} \] Thus, \[ n_{\mathrm{HCl,total}}=0.40+0.16 \] \[ n_{\mathrm{HCl,total}}=0.56\,\mathrm{mol} \] \(\textbf{Checking the dissolved magnesium inventory:}\) Both starting substances ultimately form magnesium chloride. Its total amount is \[ n_{\mathrm{MgCl_2}}=0.20+0.08 \] \[ n_{\mathrm{MgCl_2}}=0.28\,\mathrm{mol} \] The required chloride-ion amount is \(2(0.28)=0.56\,\mathrm{mol}\), matching the amount supplied by consumed HCl. The gas measurement selectively identifies the metallic component. The mass balance identifies the oxide component, while both reactions contribute to total acid consumption. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{HCl}}=0.56\,\mathrm{mol},\quad\%\mathrm{MgO}=40\%} \]

Explanation: \(\textbf{Using the inert gas as a pressure reference:}\) Initially, the only gas is argon. The solid ammonium chloride does not contribute to the ideal-gas pressure. Therefore, \[ n_{\mathrm{gas},i}=0.10\,\mathrm{mol} \] The final pressure is five times the initial pressure under unchanged temperature-volume conditions. \[ \frac{P_f}{P_i}=\frac{n_{\mathrm{gas},f}}{n_{\mathrm{gas},i}}=5 \] Thus, \[ n_{\mathrm{gas},f}=5(0.10) \] \[ n_{\mathrm{gas},f}=0.50\,\mathrm{mol} \] \(\textbf{Determining the gas newly produced:}\) Argon remains unchanged at \(0.10\,\mathrm{mol}\). Consequently, the decomposition gases together amount to \[ n_{\mathrm{NH_3}}+n_{\mathrm{HCl}}=0.50-0.10 \] \[ n_{\mathrm{NH_3}}+n_{\mathrm{HCl}}=0.40\,\mathrm{mol} \] \(\textbf{Relating gas formation to decomposed solid:}\) One mole of ammonium chloride forms one mole of ammonia and one mole of HCl. Let \(x\) be the amount decomposed. Then, \[ n_{\mathrm{NH_3}}=x \] \[ n_{\mathrm{HCl}}=x \] Therefore, \[ 2x=0.40 \] \[ x=0.20\,\mathrm{mol} \] \(\textbf{Calculating the decomposition percentage:}\) The initial ammonium chloride amount was \(0.20\,\mathrm{mol}\). \[ \%\text{decomposed}=\frac{0.20}{0.20}\times100 \] \[ \%\text{decomposed}=100\% \] \(\textbf{Calculating the ammonia mole fraction:}\) \[ X_{\mathrm{NH_3}}=\frac{0.20}{0.50} \] \[ X_{\mathrm{NH_3}}=0.40 \] \(\textbf{Checking the limiting physical condition:}\) The pressure measurement requires decomposition of exactly the entire original solid amount. Any smaller decomposition extent would produce fewer than \(0.40\,\mathrm{mol}\) of new gaseous molecules. The initial inert gas and both gaseous decomposition products must be included when calculating the final mole fraction. The pressure rise does not imply creation of matter; solid material has entered the gas phase as two molecular species. \(\textbf{Final answer:}\) \[ \boxed{\%\text{decomposed}=100\%,\quad X_{\mathrm{NH_3}}=0.40} \]

Explanation: \(\textbf{Expressing the stoichiometric feed on a mole basis:}\) One mole of CO reacts with two moles of hydrogen. Let \(x\) be the initial amount of carbon monoxide. Then, \[ n_{\mathrm{CO}}=x \] \[ n_{\mathrm{H_2}}=2x \] \(\textbf{Converting the mole ratio into a mass equation:}\) The molar masses are \[ M_{\mathrm{CO}}=28\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{H_2}}=2\,\mathrm{g\,mol^{-1}} \] The combined mass is 12.0 g. Therefore, \[ 28x+2(2x)=12.0 \] \[ 28x+4x=12.0 \] \[ 32x=12.0 \] \[ x=0.375\,\mathrm{mol} \] \(\textbf{Determining hydrogen amount and mass:}\) \[ n_{\mathrm{H_2}}=2(0.375) \] \[ n_{\mathrm{H_2}}=0.750\,\mathrm{mol} \] The hydrogen mass is \[ m_{\mathrm{H_2}}=0.750(2) \] \[ m_{\mathrm{H_2}}=1.5\,\text{g} \] \(\textbf{Following carbon into methanol:}\) Every carbon monoxide molecule contains one carbon atom. Each methanol molecule also contains one carbon atom. The balanced equation gives one mole of methanol per mole of CO consumed. Thus, \[ n_{\mathrm{CH_3OH}}=0.375\,\mathrm{mol} \] Consequently, \[ n_{C,\mathrm{methanol}}=0.375\,\mathrm{mol} \] \(\textbf{Mass-conservation verification:}\) The CO mass is \[ m_{\mathrm{CO}}=0.375(28)=10.5\,\text{g} \] Adding hydrogen gives \[ m_{\mathrm{reactants}}=10.5+1.5 \] \[ m_{\mathrm{reactants}}=12.0\,\text{g} \] The methanol mass is \[ m_{\mathrm{CH_3OH}}=0.375(32) \] \[ m_{\mathrm{CH_3OH}}=12.0\,\text{g} \] The product mass agrees with the initial reacting-gas mass. The carbon-atom amount equals the methanol molecular amount because the product contains one carbon atom per molecule. The hydrogen molecular amount, by contrast, is twice the methanol molecular amount because two hydrogen molecules react per product molecule. \(\textbf{Final answer:}\) \[ \boxed{m_{\mathrm{H_2}}=1.5\,\text{g},\quad n_{C,\mathrm{methanol}}=0.375\,\mathrm{mol}} \]

Explanation: \(\textbf{Converting pressure into a total gas amount:}\) Initially, only the inert nitrogen contributes gaseous particles. \[ n_{\mathrm{gas},i}=0.10\,\mathrm{mol} \] The final pressure is three times the initial pressure at unchanged temperature and volume. Thus, \[ n_{\mathrm{gas},f}=3(0.10) \] \[ n_{\mathrm{gas},f}=0.30\,\mathrm{mol} \] \(\textbf{Finding the new gaseous product amount:}\) Nitrogen remains unchanged. The total amount of decomposition gases is therefore \[ n_{\mathrm{CO_2}}+n_{\mathrm{H_2O(g)}}=0.30-0.10 \] \[ n_{\mathrm{CO_2}}+n_{\mathrm{H_2O(g)}}=0.20\,\mathrm{mol} \] \(\textbf{Using the equal gaseous product coefficients:}\) Two moles of sodium hydrogen carbonate produce one mole of carbon dioxide and one mole of water vapour. Let \(x\) be the amount of carbonate formed. Then, \[ n_{\mathrm{CO_2}}=x \] \[ n_{\mathrm{H_2O(g)}}=x \] Therefore, \[ 2x=0.20 \] \[ x=0.10\,\mathrm{mol} \] \(\textbf{Recovering the decomposed reactant amount:}\) Two moles of sodium hydrogen carbonate are consumed per mole of sodium carbonate formed. \[ n_{\mathrm{NaHCO_3,consumed}}=2(0.10) \] \[ n_{\mathrm{NaHCO_3,consumed}}=0.20\,\mathrm{mol} \] The initial amount was \(0.30\,\mathrm{mol}\). Therefore, \[ \%\text{decomposed}=\frac{0.20}{0.30}\times100 \] \[ \%\text{decomposed}\approx66.67\% \] \(\textbf{Determining the gas composition:}\) The gaseous mixture contains \(0.10\,\mathrm{mol}\) nitrogen, \(0.10\,\mathrm{mol}\) carbon dioxide and \(0.10\,\mathrm{mol}\) water vapour. Its total amount is \(0.30\,\mathrm{mol}\). Thus, \[ X_{\mathrm{CO_2}}=\frac{0.10}{0.30} \] \[ X_{\mathrm{CO_2}}=\frac13 \] \(\textbf{Checking the residual solid inventory:}\) The unreacted sodium hydrogen carbonate amount is \[ n_{\mathrm{NaHCO_3,remaining}}=0.30-0.20 \] \[ n_{\mathrm{NaHCO_3,remaining}}=0.10\,\mathrm{mol} \] The sodium carbonate amount is also \(0.10\,\mathrm{mol}\). Both solids remain in the vessel but do not contribute to its ideal-gas pressure. The distinction between gaseous products and solid reactant is essential to interpreting the pressure change. \(\textbf{Final answer:}\) \[ \boxed{\%\text{decomposed}=66.67\%,\quad X_{\mathrm{CO_2}}=\frac13} \]

Explanation: \(\textbf{Representing the unknown hydrocarbon:}\) Let the molecular formula be \(\mathrm{C}_{x}\mathrm{H}_{y}\), where \(x\) and \(y\) are positive integers. Its general balanced combustion equation is \[ \mathrm{C}_{x}\mathrm{H}_{y}+\left(x+\frac{y}{4}\right)\mathrm{O_2}\rightarrow x\mathrm{CO_2}+\frac{y}{2}\mathrm{H_2O} \] The coefficients follow from conservation of carbon, hydrogen and oxygen atoms. \(\textbf{Determining the carbon subscript:}\) At the same temperature and pressure, ideal-gas volume ratios correspond to molar ratios. Every carbon atom in the hydrocarbon produces one carbon dioxide molecule. Therefore, \[ x=\frac{V_{\mathrm{CO_2}}}{V_{\mathrm{hydrocarbon}}} \] \[ x=\frac{20.0}{10.0} \] \[ x=2 \] Thus, the hydrocarbon contains two carbon atoms per molecule. \(\textbf{Determining the hydrogen subscript:}\) The general combustion equation requires \[ x+\frac{y}{4} \] moles of oxygen per mole of hydrocarbon. Using the measured gas volumes, \[ x+\frac{y}{4}=\frac{V_{\mathrm{O_2}}}{V_{\mathrm{hydrocarbon}}} \] \[ 2+\frac{y}{4}=\frac{30.0}{10.0} \] \[ 2+\frac{y}{4}=3 \] \[ \frac{y}{4}=1 \] \[ y=4 \] Therefore, the molecular formula is \(\mathrm{C_2H_4}\). \(\textbf{Verifying the balanced combustion equation:}\) For ethene, the balanced equation is \[ \mathrm{C_2H_4+3O_2\rightarrow2CO_2+2H_2O} \] The corresponding gas-volume ratio at the common reference conditions is \[ V_{\mathrm{C_2H_4}}:V_{\mathrm{O_2}}:V_{\mathrm{CO_2}}=1:3:2 \] For 10.0 L hydrocarbon, the required oxygen volume is \[ V_{\mathrm{O_2}}=10.0(3) \] \[ V_{\mathrm{O_2}}=30.0\,\text{L} \] The carbon dioxide volume is \[ V_{\mathrm{CO_2}}=10.0(2) \] \[ V_{\mathrm{CO_2}}=20.0\,\text{L} \] Both values agree with the given measurements. \(\textbf{Interpreting water removal:}\) Condensation and removal of water leave carbon dioxide as the measured dry gaseous product. The original amount of carbon dioxide is unaffected by this separation. The carbon dioxide volume determines the carbon subscript, while oxygen consumption supplies the independent relationship required to determine the hydrogen subscript. The combined observations identify the molecular formula uniquely among the listed options. \(\textbf{Final answer:}\) \[ \boxed{\mathrm{C_2H_4}} \]

Explanation: \(\textbf{Representing the original molecular amounts:}\) Let \(x\) be the amount of \(\mathrm{N_2O_4}\) and \(y\) the amount of \(\mathrm{NO_2}\) initially present. The original total molecular amount gives \[ x+y=0.50 \qquad (1) \] \(\textbf{Using the final molecule count:}\) Each \(\mathrm{N_2O_4}\) molecule forms two \(\mathrm{NO_2}\) molecules. The original \(\mathrm{NO_2}\) molecules remain unchanged. Therefore, the total final molecular amount is \[ 2x+y=0.80 \qquad (2) \] Subtracting Equation (1) from Equation (2), \[ x=0.80-0.50 \] \[ x=0.30\,\mathrm{mol} \] Thus, \[ y=0.50-0.30 \] \[ y=0.20\,\mathrm{mol} \] \(\textbf{Calculating the initial dimer percentage:}\) \[ \%\mathrm{N_2O_4}=\frac{0.30}{0.50}\times100 \] \[ \%\mathrm{N_2O_4}=60\% \] \(\textbf{Comparing initial and final masses:}\) The molecular masses are \[ M_{\mathrm{NO_2}}=46\,\mathrm{g\,mol^{-1}} \] \[ M_{\mathrm{N_2O_4}}=92\,\mathrm{g\,mol^{-1}} \] Initially, \[ m_i=0.30(92)+0.20(46) \] \[ m_i=27.6+9.2 \] \[ m_i=36.8\,\text{g} \] After complete dissociation, \[ n_{\mathrm{NO_2},f}=2(0.30)+0.20 \] \[ n_{\mathrm{NO_2},f}=0.80\,\mathrm{mol} \] The final mass is \[ m_f=0.80(46) \] \[ m_f=36.8\,\text{g} \] Thus, \[ m_i:m_f=36.8:36.8 \] \[ m_i:m_f=1:1 \] \(\textbf{Interpreting the difference between mass and molecule count:}\) Dissociation increases the number of gas molecules from 0.50 to \(0.80\,\mathrm{mol}\). However, it does not create additional nitrogen or oxygen atoms. Each \(\mathrm{N_2O_4}\) molecule simply separates into two smaller nitrogen dioxide molecules. The greater molecular count is therefore compatible with exactly unchanged total mass. \(\textbf{Final answer:}\) \[ \boxed{\%\mathrm{N_2O_4}=60\%,\quad m_i:m_f=1:1} \]

Explanation: \(\textbf{Representing the two oxide amounts:}\) Let \(x\) be the amount of MO formula units and \(y\) the amount of \(\mathrm{M_2O_3}\) formula units. These are different oxides containing different numbers of metal and oxygen atoms per formula unit. \(\textbf{Applying conservation of metal atoms:}\) Every MO formula unit contains one M atom. Every \(\mathrm{M_2O_3}\) formula unit contains two M atoms. All original metal atoms are incorporated into these oxides. Therefore, \[ x+2y=0.40 \qquad (1) \] \(\textbf{Applying the independent oxygen inventory:}\) MO contains one oxygen atom per formula unit. The second oxide contains three oxygen atoms per formula unit. The measured total oxygen-atom amount gives \[ x+3y=0.45 \qquad (2) \] \(\textbf{Solving the independent atom inventories:}\) Subtracting Equation (1) from Equation (2) gives \[ y=0.45-0.40 \] \[ y=0.05\,\mathrm{mol} \] Substitute this amount into the metal-atom equation: \[ x+2(0.05)=0.40 \] \[ x=0.30\,\mathrm{mol} \] \(\textbf{Checking the two conservation constraints:}\) The reconstructed metal-atom amount is \[ 0.30+2(0.05)=0.40\,\mathrm{mol} \] The reconstructed oxygen-atom amount is \[ 0.30+3(0.05)=0.45\,\mathrm{mol} \] Both results agree with the specified atom inventories. The two calculated oxide amounts are nonnegative, and the separate balances fix them uniquely. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{MO}}=0.30\,\mathrm{mol},\quad n_{\mathrm{M_2O_3}}=0.05\,\mathrm{mol}} \]

Explanation: \(\textbf{Recovering carbon monoxide from the product inventory:}\) The balanced equation forms two moles of carbon dioxide from two moles of carbon monoxide. Thus, the consumed carbon monoxide amount equals the carbon dioxide formed. \[ n_{\mathrm{CO,consumed}}=0.40\,\mathrm{mol} \] All carbon monoxide is stated to have reacted. Therefore, \[ n_{\mathrm{CO,initial}}=0.40\,\mathrm{mol} \] \(\textbf{Recovering the oxygen consumed:}\) Two moles of carbon monoxide consume one mole of oxygen gas. \[ n_{\mathrm{O_2,consumed}}=0.40\left(\frac12\right) \] \[ n_{\mathrm{O_2,consumed}}=0.20\,\mathrm{mol} \] \(\textbf{Reconstructing the original oxygen supply:}\) The final mixture contains \(0.10\,\mathrm{mol}\) unreacted oxygen. Therefore, \[ n_{\mathrm{O_2,initial}}=0.20+0.10 \] \[ n_{\mathrm{O_2,initial}}=0.30\,\mathrm{mol} \] \(\textbf{Counting the initial gaseous amount:}\) The original gases are carbon monoxide, oxygen and inert gas. \[ n_i=0.40+0.30+0.20 \] \[ n_i=0.90\,\mathrm{mol} \] \(\textbf{Counting the final gaseous amount:}\) The final gases are carbon dioxide, excess oxygen and inert gas. \[ n_f=0.40+0.10+0.20 \] \[ n_f=0.70\,\mathrm{mol} \] \(\textbf{Calculating the pressure ratio:}\) For ideal gases in the rigid vessel at constant temperature, \[ \frac{P_f}{P_i}=\frac{n_f}{n_i} \] \[ \frac{P_f}{P_i}=\frac{0.70}{0.90} \] \[ \frac{P_f}{P_i}=\frac79 \] \(\textbf{Checking the chemical constraint:}\) The initial oxygen amount of \(0.30\,\mathrm{mol}\) exceeds the \(0.20\,\mathrm{mol}\) required for complete oxidation of CO. It is therefore consistent for all CO to disappear while oxygen remains. The inert gas contributes equally to both initial and final amounts but does not influence the chemical coefficient ratio. The final gas analysis permits reconstruction of the initial reagent amounts before the pressure comparison is made. \(\textbf{Final answer:}\) \[ \boxed{n_{\mathrm{O_2,initial}}=0.30\,\mathrm{mol},\quad\frac{P_f}{P_i}=\frac79} \]

Explanation: \(\textbf{Recovering the initial iron oxide amount:}\) The molar mass of iron(III) oxide is \[ M_{\mathrm{Fe_2O_3}}=2(56)+3(16) \] \[ M_{\mathrm{Fe_2O_3}}=160\,\mathrm{g\,mol^{-1}} \] Therefore, \[ n_{\mathrm{Fe_2O_3,initial}}=\frac{32.0}{160} \] \[ n_{\mathrm{Fe_2O_3,initial}}=0.20\,\mathrm{mol} \] \(\textbf{Using carbon dioxide to measure actual reduction:}\) Each mole of iron(III) oxide consumed produces three moles of carbon dioxide. The observed carbon dioxide amount is \(0.30\,\mathrm{mol}\). Therefore, \[ n_{\mathrm{Fe_2O_3,consumed}}=\frac{0.30}{3} \] \[ n_{\mathrm{Fe_2O_3,consumed}}=0.10\,\mathrm{mol} \] \(\textbf{Determining the reduced percentage:}\) \[ \%\text{reduced}=\frac{0.10}{0.20}\times100 \] \[ \%\text{reduced}=50\% \] \(\textbf{Recovering the unreacted iron oxide:}\) \[ n_{\mathrm{Fe_2O_3,remaining}}=0.20-0.10 \] \[ n_{\mathrm{Fe_2O_3,remaining}}=0.10\,\mathrm{mol} \] The remaining oxide mass is \[ m_{\mathrm{Fe_2O_3,remaining}}=0.10(160) \] \[ m_{\mathrm{Fe_2O_3,remaining}}=16.0\,\text{g} \] \(\textbf{Calculating the elemental iron formed:}\) One mole of oxide produces two moles of iron. \[ n_{\mathrm{Fe,formed}}=2(0.10) \] \[ n_{\mathrm{Fe,formed}}=0.20\,\mathrm{mol} \] The iron mass is \[ m_{\mathrm{Fe}}=0.20(56) \] \[ m_{\mathrm{Fe}}=11.2\,\text{g} \] \(\textbf{Determining the total retained solid mass:}\) The solids comprise the unreacted oxide and newly formed iron. \[ m_{\mathrm{solid,final}}=16.0+11.2 \] \[ m_{\mathrm{solid,final}}=27.2\,\text{g} \] \(\textbf{Checking the transfer of oxygen:}\) The initial oxide weighed 32.0 g, while final solids weigh 27.2 g. The loss of solid mass is \[ \Delta m_{\mathrm{solid}}=32.0-27.2 \] \[ \Delta m_{\mathrm{solid}}=4.8\,\text{g} \] The reacted \(0.10\,\mathrm{mol}\) oxide initially contained \[ n_{O,\mathrm{removed}}=3(0.10)=0.30\,\mathrm{mol} \] of oxygen atoms. Their mass is \[ m_{O,\mathrm{removed}}=0.30(16)=4.8\,\text{g} \] The missing solid-phase oxygen becomes part of the gaseous carbon dioxide. This independently confirms the solid-mass result. \(\textbf{Final answer:}\) \[ \boxed{\%\text{reduced}=50\%,\quad m_{\mathrm{solid,final}}=27.2\,\text{g}} \]

Explanation: Statement I is correct because decomposition produces solid calcium oxide and gaseous carbon dioxide. The gas escapes from the open vessel, so the mass of the retained solid is lower than that of the original calcium carbonate. For every 100 g of completely decomposed pure calcium carbonate, 56 g of calcium oxide remains and 44 g of carbon dioxide escapes. Statement II is correct because calcium remains within the solid phase as calcium oxide under the specified reaction. Each original carbonate formula unit contains one calcium atom, and each oxide formula unit also contains one. No calcium-containing solid is lost, so the calcium-atom inventory within the vessel remains unchanged. Statement III is incorrect because the escaping species is carbon dioxide, not elemental oxygen alone. The gas contains both carbon and oxygen. Its mass is therefore the combined mass of the carbon and two oxygen atoms transferred out of the solid per decomposed formula unit. For one mole of calcium carbonate, the mass lost is 44 g. The original carbonate contains 48 g oxygen, so these masses are not equal. Statement IV is correct because collecting the escaped carbon dioxide restores the complete material inventory for the reaction. The retained calcium oxide and collected carbon dioxide together contain all the original calcium, carbon and oxygen atoms. Their masses consequently sum to the initial carbonate mass. The open vessel loses mass because a gaseous product escapes; the reaction itself does not violate mass conservation. The key distinction is between the material remaining inside an open apparatus and the complete set of reaction products. Statements I, II and IV are correct.

Explanation: \(\textbf{Recovering magnesium carbonate from retained magnesium:}\) The magnesium atoms in the final oxide must have originated from magnesium carbonate. Each magnesium carbonate formula unit contains one magnesium atom. The measured magnesium mass is 2.4 g. Therefore, \[ n_{\mathrm{Mg}}=\frac{2.4}{24} \] \[ n_{\mathrm{Mg}}=0.10\,\mathrm{mol} \] Thus, the original magnesium carbonate amount was \[ n_{\mathrm{MgCO_3}}=0.10\,\mathrm{mol} \] \(\textbf{Calculating its original mass:}\) The magnesium carbonate molar mass is \[ M_{\mathrm{MgCO_3}}=24+12+3(16) \] \[ M_{\mathrm{MgCO_3}}=84\,\mathrm{g\,mol^{-1}} \] Therefore, \[ m_{\mathrm{MgCO_3}}=0.10(84) \] \[ m_{\mathrm{MgCO_3}}=8.4\,\text{g} \] Its initial mass percentage is \[ \%\mathrm{MgCO_3}=\frac{8.4}{15.0}\times100 \] \[ \%\mathrm{MgCO_3}=56\% \] \(\textbf{Recovering calcium carbonate from the total mass:}\) The calcium carbonate mass must have been \[ m_{\mathrm{CaCO_3}}=15.0-8.4 \] \[ m_{\mathrm{CaCO_3}}=6.6\,\text{g} \] Its amount was \[ n_{\mathrm{CaCO_3}}=\frac{6.6}{100} \] \[ n_{\mathrm{CaCO_3}}=0.066\,\mathrm{mol} \] \(\textbf{Predicting carbon dioxide from complete decomposition:}\) Both carbonates produce one mole of carbon dioxide per mole of formula units. The predicted carbon dioxide amount is \[ n_{\mathrm{CO_2,predicted}}=0.10+0.066 \] \[ n_{\mathrm{CO_2,predicted}}=0.166\,\mathrm{mol} \] Its mass is \[ m_{\mathrm{CO_2,predicted}}=0.166(44) \] \[ m_{\mathrm{CO_2,predicted}}=7.304\,\text{g} \] \(\textbf{Testing the independent gas measurement:}\) The observed carbon dioxide mass is only 6.6 g. Therefore, \[ m_{\mathrm{CO_2,observed}}\neq m_{\mathrm{CO_2,predicted}} \] The discrepancy is \[ 7.304-6.600=0.704\,\text{g} \] \(\textbf{Interpreting the inconsistency:}\) The magnesium measurement and original total mass determine the initial carbonate composition uniquely. That composition predicts more carbon dioxide than the amount reported. Because decomposition is stated to be complete, the carbonate samples are pure and no competing reaction is specified, the difference cannot be explained by an allowed alternative product. The inferred magnesium carbonate mass percentage is 56%, but the complete set of measurements is inconsistent. A strong consistency test must check whether independently measured quantities reproduce the same stoichiometric composition. \(\textbf{Final answer:}\) \[ \boxed{\%\mathrm{MgCO_3}=56\%,\quad m_{\mathrm{CO_2,predicted}}=7.304\,\text{g}\neq6.6\,\text{g}} \]

Explanation: The balanced equation specifies a fixed relationship among the four participating amounts. For every two moles of A consumed, three moles of B are consumed, two moles of C form and one mole of D forms. The corresponding reaction-amount ratio is \[ n_A:n_B:n_C:n_D=2:3:2:1 \] A table entry must satisfy all parts of this relationship, not only the reactant consumption ratio. For P, \(0.20\,\mathrm{mol}\) A requires \(0.30\,\mathrm{mol}\) B and produces \(0.20\,\mathrm{mol}\) C and \(0.10\,\mathrm{mol}\) D. All four values agree with the equation. For Q, \(0.40\,\mathrm{mol}\) A corresponds to twice the reaction extent of P. The predicted amounts are 0.60 mol B, 0.40 mol C and 0.20 mol D, matching the record. For R, \(0.30\,\mathrm{mol}\) A corresponds to \(0.15\,\mathrm{mol}\) of stoichiometric reaction units. The predicted B consumption is \[ n_B=0.30\left(\frac32\right)=0.45\,\mathrm{mol} \] The predicted C production is \[ n_C=0.30\left(\frac22\right)=0.30\,\mathrm{mol} \] However, the predicted D amount is \[ n_D=0.30\left(\frac12\right)=0.15\,\mathrm{mol} \] The table instead reports \(0.10\,\mathrm{mol}\) D. Trial R is therefore inconsistent with the stated reaction. For S, \(0.50\,\mathrm{mol}\) A requires \(0.75\,\mathrm{mol}\) B and forms \(0.50\,\mathrm{mol}\) C and \(0.25\,\mathrm{mol}\) D. All of its entries agree. Stoichiometric coefficients express proportions, so measured reacting amounts need not be whole-number mole values. The correct conclusion is that P, Q and S are consistent, while R is not.

Explanation: \(\textbf{Recovering the empirical composition from constituent atoms:}\) The given atomic amounts are \[ n_C=0.10\,\mathrm{mol} \] \[ n_O=0.20\,\mathrm{mol} \] Their simplest ratio is \[ C:O=0.10:0.20 \] \[ C:O=1:2 \] The empirical formula is therefore \(\mathrm{CO_2}\). \(\textbf{Checking the sample mass against its elemental inventory:}\) The carbon contributes \[ m_C=0.10(12) \] \[ m_C=1.2\,\text{g} \] The oxygen contributes \[ m_O=0.20(16) \] \[ m_O=3.2\,\text{g} \] The total is \[ m_C+m_O=1.2+3.2 \] \[ m_C+m_O=4.4\,\text{g} \] Thus, the sample mass agrees with its measured atomic amounts. \(\textbf{Interpreting the independent molecular-entity amount:}\) The second experiment claims that 4.4 g contains \(0.10\,\mathrm{mol}\) molecular entities. This corresponds to a molar mass of \[ M_{\mathrm{molecular}}=\frac{4.4}{0.10} \] \[ M_{\mathrm{molecular}}=44\,\mathrm{g\,mol^{-1}} \] \(\textbf{Testing the molecular multiplier:}\) The empirical formula mass of carbon dioxide is \[ M_{\mathrm{CO_2}}=12+2(16) \] \[ M_{\mathrm{CO_2}}=44\,\mathrm{g\,mol^{-1}} \] The molecular multiplier is \[ k=\frac{44}{44} \] \[ k=1 \] The molecular formula is therefore also \(\mathrm{CO_2}\). \(\textbf{Final consistency check:}\) The independent molecular measurement agrees with both the empirical formula and the elemental mass inventory. It is therefore consistent, not inconsistent. The independent molecular measurement confirms the empirical formula rather than contradicting it. The three independent observations mutually reinforce the molecular formula \(\mathrm{CO_2}\). \(\textbf{Final answer:}\) \[ \mathrm{CO_2},\quad M=44\,\mathrm{g\,mol^{-1}} \]